小于等于给定整数的 2 的最高幂
给定一个整数 N ，任务是找到小于或等于 N 的 2 的最高幂。
示例:
输入: N = 9 输出: 8 说明: 2 小于 9 的最高幂是 8。
输入: N = -20 输出: -32 说明: 2 小于-20 的最高功率为-32。
输入: N = -84 输出: -128
方法:思路是用对数来解决上述问题。对于任何给定的数字 N，它可以是正的，也可以是负的。对于每种情况，可以执行以下任务:
- 如果输入为正:楼层(原木 2 (N)) 即可计算。
- 如果输入为负:可以计算 ceil(log 2 (N)) ，并在该值上加上一个 -ve 符号。
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the lowest power
// of 2 close to given positive number
int powOfPositive(int n)
{
// Floor function is used to determine
// the value close to the number
int pos = floor(log2(n));
return pow(2, pos);
}
// Function to return the lowest power
// of 2 close to given negative number
int powOfNegative(int n)
{
// Ceil function is used for negative numbers
// as -1 > -4\. It would be opposite
// to positive numbers where 1 < 4
int pos = ceil(log2(n));
return (-1 * pow(2, pos));
}
// Function to find the highest power of 2
void highestPowerOf2(int n)
{
// To check if the given number
// is positive or negative
if (n > 0) {
cout << powOfPositive(n);
}
else {
// If the number is negative,
// then the ceil of the positive number
// is calculated and
// negative sign is added
n = -n;
cout << powOfNegative(n);
}
}
// Driver code
int main()
{
int n = -24;
highestPowerOf2(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
class GFG
{
// Function to return the lowest power
// of 2 close to given positive number
static int powOfPositive(int n)
{
// Floor function is used to determine
// the value close to the number
int pos = (int)Math.floor((Math.log(n)/Math.log(2)));
return (int)Math.pow(2, pos);
}
// Function to return the lowest power
// of 2 close to given negative number
static int powOfNegative(int n)
{
// Ceil function is used for negative numbers
// as -1 > -4\. It would be opposite
// to positive numbers where 1 < 4
int pos = (int)Math.ceil((Math.log(n)/Math.log(2)));
return (int)(-1 * Math.pow(2, pos));
}
// Function to find the highest power of 2
static void highestPowerOf2(int n)
{
// To check if the given number
// is positive or negative
if (n > 0)
{
System.out.println(powOfPositive(n));
}
else
{
// If the number is negative,
// then the ceil of the positive number
// is calculated and
// negative sign is added
n = -n;
System.out.println(powOfNegative(n));
}
}
// Driver code
public static void main (String[] args)
{
int n = -24;
highestPowerOf2(n);
}
}
// This code is contributed by AnkitRai01
C
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the lowest power
// of 2 close to given positive number
static int powOfPositive(int n)
{
// Floor function is used to determine
// the value close to the number
int pos = (int)Math.Floor((Math.Log(n)/Math.Log(2)));
return (int)Math.Pow(2, pos);
}
// Function to return the lowest power
// of 2 close to given negative number
static int powOfNegative(int n)
{
// Ceil function is used for negative numbers
// as -1 > -4\. It would be opposite
// to positive numbers where 1 < 4
int pos = (int)Math.Ceiling((Math.Log(n)/Math.Log(2)));
return (int)(-1 * Math.Pow(2, pos));
}
// Function to find the highest power of 2
static void highestPowerOf2(int n)
{
// To check if the given number
// is positive or negative
if (n > 0)
{
Console.WriteLine(powOfPositive(n));
}
else
{
// If the number is negative,
// then the ceil of the positive number
// is calculated and
// negative sign is added
n = -n;
Console.WriteLine(powOfNegative(n));
}
}
// Driver code
public static void Main()
{
int n = -24;
highestPowerOf2(n);
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the above approach
from math import floor,ceil,log2
# Function to return the lowest power
# of 2 close to given positive number
def powOfPositive(n) :
# Floor function is used to determine
# the value close to the number
pos = floor(log2(n));
return 2**pos;
# Function to return the lowest power
# of 2 close to given negative number
def powOfNegative(n) :
# Ceil function is used for negative numbers
# as -1 > -4\. It would be opposite
# to positive numbers where 1 < 4
pos = ceil(log2(n));
return (-1 * pow(2, pos));
# Function to find the highest power of 2
def highestPowerOf2(n) :
# To check if the given number
# is positive or negative
if (n > 0) :
print(powOfPositive(n));
else :
# If the number is negative,
# then the ceil of the positive number
# is calculated and
# negative sign is added
n = -n;
print(powOfNegative(n));
# Driver code
if __name__ == "__main__" :
n = -24;
highestPowerOf2(n);
# This code is contributed by AnkitRai01
java 描述语言
<script>
// Javascript implementation of the above approach
// Function to return the lowest power
// of 2 close to given positive number
function powOfPositive(n)
{
// Floor function is used to determine
// the value close to the number
let pos = Math.floor(Math.log2(n));
return Math.pow(2, pos);
}
// Function to return the lowest power
// of 2 close to given negative number
function powOfNegative(n)
{
// Ceil function is used for negative numbers
// as -1 > -4\. It would be opposite
// to positive numbers where 1 < 4
let pos = Math.ceil(Math.log2(n));
return (-1 * Math.pow(2, pos));
}
// Function to find the highest power of 2
function highestPowerOf2(n)
{
// To check if the given number
// is positive or negative
if (n > 0)
{
document.write(powOfPositive(n));
}
else
{
// If the number is negative,
// then the ceil of the positive number
// is calculated and
// negative sign is added
n = -n;
document.write(powOfNegative(n));
}
}
// Driver code
let n = -24;
highestPowerOf2(n);
// This code is contributed by mohit kumar 29
</script>
Output:
-32
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